How do you solve the following system: #2x-4y=6 , y= 3/4x+3 2 #?

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Answer 1 · 2016-02-27T01:24:15

x=-134; y=137/2

#a) 2x-4y=6#

#b) y=3/4x+32#

You just need to replace #y# in #a)# by its value in #b#:

#a) 2x-4(3/4x+32)=6#

#a) 2x-3x-128=6#

#a) -x=6+128=134#

#a) x=-134#

Now that we have the value of x we wil replace it in equation b:

#b) y=(3/4)xx(-134)+32=-201/2+32= 137/2#